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Tutor profile: Shari R.

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Shari R.
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Questions

Subject:German

TutorMe
Question:

Change the following present tense sentence to past tense: Ich habe viele Freunde und spiele gerne mit ihnen Fußball.

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Shari R.

Ich hatte viele Freunde und habe gern mit ihnen Fußball gespielt.

Subject:Calculus

TutorMe
Question:

Given the following function: $$f(x) = 6x^{2} + 10$$ a) Find the average rate of change from the interval [0,10] b) Find the instantaneous rate of change at $$x = 5$$

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Shari R.

Step 1: Both part a and b are asking for the rate of change of the function which means the derivative of the function. Take the derivative: $$f'(x) = 12x$$ Step 2: For part a, use the formula for rate of change over an interval: $$f'(x) = f'(b) - f'(a), [a,b]$$ $$f'(x) = 12(10) - 12(0), [0,10]$$ $$f'(x) = 120$$ Step 3: For part b, plug $$x = 5$$ into $$f'(x) = 12x$$: $$f'(5) = 12(5)$$ $$f'(5) = 60$$

Subject:Algebra

TutorMe
Question:

Solve the following quadratic equation using the quadratic formula: $$5x^{2} + 6x = -1$$

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Shari R.

Step 1: Put the given equation into the form of $$ax^{2} + b{x} + c = 0$$: $$5x^{2} + 6x + 1 = 0$$ Step 2: Identify a, b, and c: $$a = 5, b = 6, c = 1$$ Step 3: Write down the quadratic equation: $$x = \frac{-b \pm \sqrt{b^{2} - 4(a)(c)}}{2(a)}$$ Step 4: Plug values from Step 2 into the quadratic equation: $$x = \frac{-6 \pm \sqrt{6^{2} - 4(5)(1)}}{2(5)}$$ Step 5: Solve: $$x = \frac{-6 \pm \sqrt{36 - 20}}{10}$$ $$x = \frac{-6 \pm \sqrt{16}}{10}$$ $$x = \frac{-6 \pm 4}{10}$$ Solve for each root: $$x = \frac{-6 + 4}{10}$$ $$x = \frac{-6 - 4}{10}$$ $$x = \frac{-1}{5}$$ $$x = -1$$ Step 6: Check your answer by plugging it in to the original equation: Solving for $$x = \frac{-1}{5}$$: $$5x^{2} + 6x + 1 = 0$$ $$5(-\frac{1}{5})^{2} + 6(-\frac{1}{5}) + 1 = 0$$ $$\frac{5}{25} + -\frac{6}{5} + 1 = 0$$ $$\frac{1}{5} + -\frac{6}{5} + 1 = 0$$ $$0= 0$$ Solving for $$x = -1$$: $$5x^{2} + 6x + 1 = 0$$ $$5(-1)^{2} + 6(-1) + 1 = 0$$ $$5 - 6 + 1 = 0$$ $$0= 0$$

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